Bourbaki

Proposition 4

Tag 04N5, permanent. Cite it as [Bourbaki, Tag 04N5].

Commutative Algebra, chapter V, § 1. Notion of an integral element

Let AA be a ring, RR an AA-algebra (not necessarily commutative) and (xi)1≤i≤n(x_i)_{1 \leq i \leq n} a finite family of elements of RR which are pairwise permutable. If, for all ii, xix_i is integral over A[x1,…,xi−1]A[x_1, \ldots, x_{i-1}] (and in particular if all the xix_i are integral over AA), then the subalgebra A[x1,…,xn]A[x_1, \ldots, x_n] of RR is a finitely generated AA-module.

We argue by induction on nn, the proposition being just (E_{II}) for n=1n = 1. The induction hypothesis implies that B=A[x1,…,xn−1]B = A[x_1, \ldots, x_{n-1}] is a finitely generated A-module; as xnx_n is integral over BB, B[xn]=A[x1,…,xn]B[x_n] = A[x_1, \ldots, x_n] is a finitely generated B-module and therefore also a finitely generated A-module (Algebra, Chapter II, § 1, no. 13, Proposition 25).

Bourbaki puts the proof directly after the statement, so what is above is the statement and its proof together.

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Transcribed from the printed volume. Not translated.